Ch. 21: Gauss’s Law
•  Gauss’s law is an alternative description of Coulomb’s
law that allows for an easier method of determining the
electric field for situations where the charge distribution
is contains symmetry.
Flux
•  Gauss’s law relates something called electric flux with a
charge distribution. The concept of electric flux is tricky
so first let’s consider an analogy.
•  Consider a flowing river, and placing a wire rectangle in
the water. We can ask “How much water flows through
the rectangle each second?”
•  This is the same as asking “What is the flux of water through
the rectangle?”
•  The answer depends on:
•  density of water
•  velocity flow of water
•  orientation of the rectangle.
•  If rectangle is perpendicular to velocity:
Flux = Φ = ρvA
Flux
•  Suppose you rotate the rectangle. Does the flux change?
•  Sure it does!
�
Φ = ρvA cos φ = ρvA⊥ = ρ�v · A
Electric flux
•  The Electric Flux is defined as the amount of an electric
field passing through a surface
•  The electric field plays the same role as
the electric field isn’t “flowing”)
ρ�v (note though that
•  For a uniform electric field and a rectangular surface:
� ·A
�
ΦE = EA cos φ = E
Clicker question
©University of Colorado, Boulder (2008)
Electric flux with curved surfaces and
nonuniform fields
•  When the surface is curved or the field is nonuniform, we calculate
the flux by dividing the surface into small patches dA, so small that
each patch is essentially flat and the field is essentially uniform
over each.
•  The flux of each patch is
�
� · dA
dΦ = E
•  The total flux is given by summing
up the contribution of each patch:
•  In the limit of infinitely many
infinitesimally small patches,
the sum becomes a
surface integral:
ΦE =
�
surface
�
� · dA
E
Gaussian Surfaces
•  A Gaussian surface is any mathematical closed surface
that fully encloses a 3D volume.
• Valid examples • Not Valid
•  We will look at the electric flux through Gaussian
surfaces.
•  By convention, electric fields going from the inside to the
outside of the surface produces a positive flux, and fields
going from the outside to the inside produces a negative
flux
•  Now let’s look at the relationship between the electric
flux through a Gaussian surface and the charge
distribution.
Clicker Question
•  A dipole moment is situated inside a Gaussian surface
having the shape of a cuboid (rectangular box). The net
electric flux through this surface is
1.  positive
2.  negative
3.  zero
4.  depends how and exactly where the dipole is situated inside
Clicker Question
Field lines due to a point-charge are shown
below, as well as a cylindrical Gaussian
surface.What is the net electric flux through
the surface?
A: 0
B: positive
C: negative
Gauss’s law
•  The electric flux through any closed surface is proportional to
the charge enclosed.
�
qenclosed
�
�
E · dA =
�0
�0 =
1
= 8.85 × 10−12 C2 /N m2
4πk
•  The contour integral is over the Gaussian surface; qenclosed is the
charge enclosed by that surface.
•  Gauss’s law is true for any surface and charge
distribution
•  Gauss’s law is equivalent to Coulomb’s law;
both describe the inverse square dependence
of the point-charge field.
•  One of the four fundamental laws of electromagnetism
Gauss’s Law and Point Charge
•  Gaussian shell of radius r with charge in center
•  Spherical symmetry allows one to claim E is constant (in
magnitude) on Gaussian surface
� •
�
E is also parallel to the normal everywhere on surface
2
�
�
E · dA = 4πr E
Q
E=k 2
r
1
� = 4πkQ = Q
� · dA
E
�0
This holds even if charge is not in center of sphere.
CT 27.7
Three closed surfaces (2 spheres and a cube)
enclose a + point charge.
Which surface has the largest electric flux through it?
A: Small (pink) cube
B: middle (yellow)
sphere
C: larger (green) sphere
green
yellow
pink
D: all 3 are same
E: Need more info
©University of Colorado, Boulder (2008)
+
CT 27.6b
A single charge Q is just outside a spherical
Gaussian surface.
What is the electric field
inside the Gaussian
surface?
A: 0 everywhere inside
B: non-zero everywhere
in the sphere
C: Not enough info given
©University of Colorado, Boulder (2008)
+
Using Gauss’s law
•  Gauss’s law is always true.
•  But it’s only useful for calculating the electric field in
situations with sufficient symmetry:
•  Spherical symmetry
•  Line symmetry
•  Plane symmetry
Gauss’s law is always true, so it holds in both situations
shown. Both surfaces surround the same net charge, so the
flux through each is the same. But only the left-hand
situation has enough symmetry to allow the use of Gauss’s
law to calculate the field. The electric fields differ in the
two situations, even though the fluxes are the same
CT 27.6
A spherical shell has a uniform positive charge
density on its surface. (There are no other
charges around)
+ +
+
What is the electric field
+
+
+
+
inside the sphere?
A: E=0 everywhere inside +
+
+
+
B: E is non-zero
+
everywhere in the sphere
+ + ++
C: E=0 only at the very center, but non-zero
elsewhere inside the sphere.
D: Not enough info given
©University of Colorado, Boulder (2008)
2
QA = σAA = σrA
dΩ
2
QB = σAB = σrB
dΩ
dΩ = 2π sin θ dθ
EA ∝
EB ∝
2
QA /rA
2
QB /rB
⇒ EA = EB
CT 27.16
The total charge on the thin spherical shell is Q.
(Uniformly spread out)
Which statement is true outside the spherical
shell?
!
+ + + +
Q
A) E =
rˆ
+
2
+
4 "#0 r
+
+
+
+ !
Q
+
+ B) E is not
rˆ
2
+ + +
4 "# r
!
!
©University of Colorado, Boulder (2008)
0
The field of a uniformly charged sphere
•  The charge distribution has spherical symmetry, so the appropriate
Gaussian surfaces are spheres.
•  Outside the sphere, the enclosed charge is
the total charge Q.
�
� = 4πr2 E
� · dA
Φ= E
�
qenclosed
�
�
E · dA =
�0
kQ
4π r E = Q ε 0 , so E =
= 2.
2
4πε 0 r
r
2
Q
•  Thus the field outside a spherical charge distribution is identical to
that of a point charge.
The field of a uniformly charged sphere
•  Inside the sphere:
�
qenclosed
�
�
E · dA =
�0
�
� = 4πr2 E
� · dA
E
� r �3
4π 3
qenclosed = ρ r = Q
3
R
Φ=
� r �3
1
4πr2 E = Q
�0
R
Qr
kQ r
E=
=
3
4π�0 R
R3
Cylindrical symmetry
•  The charge density depends only on the perpendicular distance
from the cylindrical axis.
•  This requires a charge distribution that is infinitely long.
•  However, cylindrical symmetry is a good approximation for finite cylindrical
charge distributions with length much greater than diameter, at points close
to the charge.
•  Example: A infinitely long line charge.
•  By symmetry, E field must
be radial
•  Want a Gaussian surface
that is perpendicular to
field
Cylindrical symmetry
•  Use a cylindrical Gaussian surface.
Φ = 2πrLE
qenclosed = λ L
�
� = qenc /�0
� · dA
E
λ
E=
2π�0 r
Plane symmetry
•  In plane symmetry, the charge density depends only on the perpendicular
distance from a plane (the plane of symmetry).
•  This requires a charge distribution that extends infinitely in two directions.
•  However, plane symmetry is a good approximation for finite slabs of charge whose
thickness is much less than their extent in the other two dimensions, at points close to
the charge.
•  By symmetry, the electric field must be perpendicular to the plane of
symmetry (say the z direction) and its magnitude can only depend on its
distance from this plane of symmetry.
Example: Charged Plane
•  Want Gaussian surface such that surface is either parallel to electric field (which
contributes zero flux) or the surface is perpendicular to the electric field – The
Gaussian Pillbox
•  Set up pillbox such that each end is the
same distance from the plane. By
symmetry each end must have the same
electric flux:
1
Φ = 2AE = σA E is uniform! Does not
�0
1
decrease with distance
E=
σ
from plane.
2�0
2.18
Which is true about |E| at points on the
imaginary dashed triangle?
(I)
A.  |E| is the same everywhere (uniform) in (I) ONLY
B.  |E| is uniform in (II) ONLY
C.  Uniform in both, but different for cases I and II
D. Uniform in both, and same for cases I and II.
E. |E| varies from point to point in both cases
(II)
Gauss’s law and conductors
•  Valence electrons in
conductors are free to
move, and they do so in
response to an electric
field.
•  If a conductor is allowed
to reach electrostatic
equilibrium, then charges
redistribute themselves to
cancel the applied field inside
the conductor.
•  Therefore the electric field is
zero inside a conductor (in
electrostatic equilibrium).
•  The electric field must be
perpendicular to the
surface (a horizontal
component would cause
charges to accelerate/
move). Very near surface,
1
E≈ σ
�0
•  Gauss’s law requires that
any free charge on a
conductor reside on the
conductor surface.
CT 27.19
A negative point charge with
charge -Q sits in the interior of a
thick spherical (conducting) neutral
metal shell. What is the total charge
on the inner surface of the shell?
Outer
surface
-Q
Gaussian
surface
Inner
surface
©University of Colorado, Boulder (2008)
A: -Q
B: +Q
C: +2Q
D: zero
E: + Q/2
A point charge +Q sits outside a hollow spherical
conducting shell. The electric field at point P inside of
the shell is
a)  zero
b)  directed to the right
c)  directed to the left
Faraday Cage
2.30b
A HOLLOW copper sphere has no net charge.
A charge, q , is in the hole, SHIFTED right a bit
from the center.
(We are in static equilibrium.)
What does the E field look like in the hole
region?
+q
A)  Simple Coulomb
field (straight away
from q , right up to
the wall)
B) Complicated/ it s
hard to compute
2.30b
A HOLLOW copper sphere has no net charge.
A charge, q , is in the hole, SHIFTED right a bit
from the center.
(We are in static equilibrium.)
What does the E field look like outside the sphere?
+q
A)  Simple Coulomb field
(radially outward from q’)
B)  Simple Coulomb field (radially
outward from center of
sphere)
B) Complicated/ it s hard to
compute
Clicker Question
Consider two parallel, large conducting plates, each
having a thickness a, oriented close to each other (this
is a parallel plate capacitor). In the figure to the right,
the plates are seen edge on. The left plate has a net
charge +Q and the right plate has a net charge –Q.
+Q
−Q
For the left plate, where does the charge reside?
1.
2.
3.
4.
All of the charge is on the right surface of the plate
All of the charge is on the left surface of the plate
Half of the charge is on each of the two plate surfaces.
The charge is distributed uniformly throughout the plate
a
Parallel Plate Capacitor
What is the electric field between the plates?
EA = σ A/�0
E = σ/�0